Tìm x biết
a\ \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
b\ \(\left(x-1\right)^2=\frac{36}{49}\)
Những câu hỏi liên quan
Tìm x
a\ \(\left(x-7\right)^{x+1}-\left(x+7\right)^{x+11}=0\)
b\ \(\left(x-1\right)^2=\frac{36}{49}\)
a,(x-7)x+1-(x-7)x+11=0
=>(x-7)x+1.[1-(x-7)10]=0
=>(x-7)x+1=0
=>x-7=0
=>x=7
hoặc 1-(x-7)10=1
=>(x-7)10=1
=>x-7=-1;1
=>x=8;6
vậy x=6;7;8
b,(x-1)2=36/49
=>x-1=6/7;-6/7
=>x=13/7;1/7
vậy x=1/7;13/7
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tim xEz biet:
a)\(x^2+\left(y-\frac{1}{4}\right)^4=6\)
b)\(x+\left(\frac{-31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x=y^2\)
c)\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
cách 1:=> (x - 7)^(x+1)= (x-7)^(x+11)
TH1: x-7=0 => x=7 => 0^8=0^18 (TM)
TH2: x-7=1 => x=8 (TM)
TH3: x khác 7 và 8 => x+1=x+11 => vô lý => loại
KL: x = 7 hoặc x=8
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( x-7)^( x+1) - ( x-7)^(x+11) = 0
( x-7)^( x+1) - ( x-7)^(x+1)*x^10 = 0
( x-7)^( x+1) (1-x^10) = 0
tới đây dễ òi
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cách 3:\(\Leftrightarrow\left(x-7\right)^{x+1}=\left(x-7\right)^{x+11}\)
\(\Leftrightarrow x-7=0\)hoặc x+1=x+11(vô lí)
\(\Rightarrow x=7\)
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1)tìm x
a) \(\text{(5x+1)}^2=\frac{36}{49}\) b)\(\left(x-2\right)^3=\left(\frac{2}{6}\right)^6\) c)\(\left(x-7\right)^{x-1}-\left(x-7\right)^{x+11}=0\)
Tìm x, biết:
a)\(3\left|x+4\right|-\left|2x+1\right|-5\left|x+3\right|+\left|x-9\right|=5\)
b)\(\left|\frac{11}{5}-x\right|+\left|x+\frac{1}{5}\right|+\frac{41}{5}=1,2\)
c)\(2\left|x+\frac{7}{2}\right|+\left|x\right|-\frac{7}{2}=\left|\frac{11}{5}-x\right|\)
Tìm x biết :
a. \(|x-\frac{1}{3}|+\frac{4}{5}=|\left(-3,2\right)+\frac{2}{5}|\)
b. \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
a. \(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\left|-\frac{16}{5}+\frac{2}{5}\right|-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\left|-\frac{14}{5}\right|-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\frac{14}{5}-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{5}{3}\end{cases}.}\)
Vậy \(x\in\left\{-\frac{5}{3};\frac{7}{3}\right\}.\)
b. \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)\(\Leftrightarrow\left(x-7\right)^{x+1}-\left(x-7\right)^{x+1}\times\left(x-7\right)^{10}=0\)\(\Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\Leftrightarrow\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}.}\)Xét 2 trường hợp:
\(\left(x-7\right)^{x+1}=0\)\(\Leftrightarrow x-7=0\Leftrightarrow x=7.\)\(1-\left(x-7\right)^{10}=0\Leftrightarrow\left(x-7\right)^{10}=1\Leftrightarrow\left(x-7\right)^{10}=\left(\pm1\right)^{10}\)\(\Leftrightarrow\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=6\end{cases}.}}\)Vậy \(x\in\left\{6;7;8\right\}.\)
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ban nguyen nhat minh giang lai cho mk dong 2 cau b cai
mk cam on
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Dòng 2 câu b là đặt nhân tử chung \(\left(x-7\right)^{x+1}\)ra ngoài
Tìm x, biết:
a)\(\left(x+5\right).\left(x+9\right)>0\)
b)\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(-\frac{1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
bạn ơi trả lời được câu này kông
( x + 1 ) + ( x - 3 ) + ( x + 5 ) + ............ + ( x +9) = 35
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tìm x:
a,\(\left(x-\frac{7}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
b,\(\left(\frac{x}{2}-5\right)^5=-32\)
c,\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
a,\(\left(x-\frac{7}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)
\(x-\frac{7}{9}=\frac{4}{9}\)
\(x=\frac{4}{9}+\frac{7}{9}\)
\(x=\frac{11}{9}\)
Vậy x=\(\frac{11}{9}\)
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Mình cần gấp bài này !!!
1/Tìm x, biết :
a/ \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}+\frac{x}{\left(x+3\right)\left(x+34\right)}\)
b/ \(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=\frac{-1}{20}\)
Tìm x biết:
a)\(\frac{2}{3}.\left(x-\frac{3}{8}\right)-x-\left(-\frac{7}{8}+\frac{2}{3}\right)=\left(\frac{-3}{4}\right)^3:1\frac{11}{16}\)
b)\(-\frac{7}{8}+\frac{7}{8}:\left(\frac{2}{3}-x\right)+\frac{5}{6}:\left(-1\frac{11}{35}\right)=\left(0,8\right)^2\)